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| author | hardythe1 | 2014-07-25 12:37:07 +0530 |
|---|---|---|
| committer | hardythe1 | 2014-07-25 12:37:07 +0530 |
| commit | efb9ead5d9758d5d0bed7a22069320b14f972e40 (patch) | |
| tree | 88c95b7c5d4bccfbcb2bdf16bf2bef0b73184808 /Electrical_Circuit_Theory_And_Technology/chapter_18-checkpoint_2.ipynb | |
| parent | 6846da7f30aadc7b3812538d1b1c9ef2e465a922 (diff) | |
| download | Python-Textbook-Companions-efb9ead5d9758d5d0bed7a22069320b14f972e40.tar.gz Python-Textbook-Companions-efb9ead5d9758d5d0bed7a22069320b14f972e40.tar.bz2 Python-Textbook-Companions-efb9ead5d9758d5d0bed7a22069320b14f972e40.zip | |
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Diffstat (limited to 'Electrical_Circuit_Theory_And_Technology/chapter_18-checkpoint_2.ipynb')
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1 files changed, 535 insertions, 0 deletions
diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_18-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_18-checkpoint_2.ipynb new file mode 100755 index 00000000..f3f727b4 --- /dev/null +++ b/Electrical_Circuit_Theory_And_Technology/chapter_18-checkpoint_2.ipynb @@ -0,0 +1,535 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:d16259f8430b68bdc5f61e2c9d378821ed7dc4c5b9f7144ffa85cfd5e017e63d"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h1>Chapter 18: Operational amplifiers</h1>"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 1, page no. 279</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Calculate the output voltage of the amplifier.\n",
+ "from __future__ import division\n",
+ "import math\n",
+ "#initializing the variables:\n",
+ "Vi2 = 2.45;# in Volts\n",
+ "Vi1 = 2.35;# in Volts\n",
+ "A0 = 120;# open-loop voltage gain\n",
+ "\n",
+ "#calculation:\n",
+ "Vo = A0*(Vi2 - Vi1)\n",
+ "\n",
+ "\n",
+ "#Results\n",
+ "print \"\\n\\n Result \\n\\n\"\n",
+ "print \"\\n the output voltage is \",round(Vo,2),\" V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "\n",
+ " Result \n",
+ "\n",
+ "\n",
+ "\n",
+ " the output voltage is 12.0 V"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 2, page no. 281</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Determine the common-mode gain of an op amp\n",
+ "from __future__ import division\n",
+ "import math\n",
+ "#initializing the variables:\n",
+ "Vg = 150E3;# differential voltage gain \n",
+ "CMRR = 90;# in dB\n",
+ "\n",
+ "#calculation:\n",
+ "CMG = Vg/(10**(CMRR/20))\n",
+ "\n",
+ "\n",
+ "#Results\n",
+ "print \"\\n\\n Result \\n\\n\"\n",
+ "print \"\\n common-mode gain is \",round(CMG,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "\n",
+ " Result \n",
+ "\n",
+ "\n",
+ "\n",
+ " common-mode gain is 4.74"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 3, page no. 282</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Calculate the common-mode gain and the CMRR.\n",
+ "from __future__ import division\n",
+ "import math\n",
+ "#initializing the variables:\n",
+ "Vg = 120;# differential voltage gain \n",
+ "Vi = 3;# in Volts\n",
+ "Vo = 0.024;# in Volts\n",
+ "\n",
+ "#calculation:\n",
+ "CMG = Vo/Vi\n",
+ "CMRR = 20*(1/2.303)*math.log(Vg/CMG)\n",
+ "\n",
+ "\n",
+ "#Results\n",
+ "print \"\\n\\n Result \\n\\n\"\n",
+ "print \"\\n common-mode gain is \",round(CMG,3),\" and CMRR is \",round(CMRR,2),\" dB\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "\n",
+ " Result \n",
+ "\n",
+ "\n",
+ "\n",
+ " common-mode gain is 0.008 and CMRR is 83.51 dB"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 4, page no. 283</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Determine the output voltage when the input voltage is: (a) +0.4 V (b) -1.2 V\n",
+ "from __future__ import division\n",
+ "import math\n",
+ "#initializing the variables:\n",
+ "Rf = 2000;# in ohms\n",
+ "Ri = 1000;# in ohms\n",
+ "Vi1 = 0.4;# in Volts\n",
+ "Vi2 = -1.2;# in Volts\n",
+ "\n",
+ "#calculation:\n",
+ "Vo1 = -1*Rf*Vi1/Ri\n",
+ "Vo2 = -1*Rf*Vi2/Ri\n",
+ "\n",
+ "\n",
+ "#Results\n",
+ "print \"\\n\\n Result \\n\\n\"\n",
+ "print \"\\n output voltage when the input voltage is 0.4V is \",round(Vo1,2),\" V \"\n",
+ "print \" and when the input voltage is -1.2V is \",round(Vo2,2),\" V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "\n",
+ " Result \n",
+ "\n",
+ "\n",
+ "\n",
+ " output voltage when the input voltage is 0.4V is -0.8 V \n",
+ " and when the input voltage is -1.2V is 2.4 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 5, page no. 283</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Calculate (a) the voltage gain, and\n",
+ "#(b) the output offset voltage due to the input bias current. \n",
+ "#(c) How can the effect of input bias current be minimised?\n",
+ "from __future__ import division\n",
+ "import math\n",
+ "#initializing the variables:\n",
+ "Ii = 100E-9;# in Amperes\n",
+ "T = 20;# in \u00b0C\n",
+ "Rf = 1E6;# in ohms\n",
+ "Ri = 10000;# in ohms\n",
+ "\n",
+ "#calculation:\n",
+ "A = -1*Rf/Ri\n",
+ "Vos = Ii*Ri*Rf/(Ri+Rf)\n",
+ "\n",
+ "\n",
+ "#Results\n",
+ "print \"\\n\\n Result \\n\\n\"\n",
+ "print \"\\n (a)the voltage gain is \",round(A,2),\"\"\n",
+ "print \"\\n (b)output offset voltage is \",round(Vos*1000,2),\" mV\"\n",
+ "print \"\\n (c)The effect of input bias current can be minimised by ensuring \"\n",
+ "print \"that both inputs have the same driving resistance.\" \n",
+ "print \"This means that a resistance of value of 9.9 kohm (from part (b)) \"\n",
+ "print \"should be placed between the non-inverting (+) terminal and earth.\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "\n",
+ " Result \n",
+ "\n",
+ "\n",
+ "\n",
+ " (a)the voltage gain is -100.0 \n",
+ "\n",
+ " (b)output offset voltage is 0.99 mV\n",
+ "\n",
+ " (c)The effect of input bias current can be minimised by ensuring \n",
+ "that both inputs have the same driving resistance.\n",
+ "This means that a resistance of value of 9.9 kohm (from part (b)) \n",
+ "should be placed between the non-inverting (+) terminal and earth.\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 6, page no. 284</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Design an inverting amplifier\n",
+ "from __future__ import division\n",
+ "import math\n",
+ "#initializing the variables:\n",
+ "Vg = 40;# in dB\n",
+ "bf = 5000;# in Hz\n",
+ "Ri = 10000;# in ohms\n",
+ "\n",
+ "#calculation:\n",
+ "A = 10**(Vg/20)\n",
+ "Rf = A*Ri\n",
+ "f = A*bf\n",
+ "\n",
+ "\n",
+ "#Results\n",
+ "print \"\\n\\n Result \\n\\n\"\n",
+ "print \"\\n the voltage gain is \",round(A,2),\", Rf = \",round(Rf/1000,2),\"kohm and frequency = \",round(f/1000,2),\" kHz\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "\n",
+ " Result \n",
+ "\n",
+ "\n",
+ "\n",
+ " the voltage gain is 100.0 , Rf = 1000.0 kohm and frequency = 500.0 kHz"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 7, page no. 286</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#determine (a) the voltage gain (b) the output voltage\n",
+ "from __future__ import division\n",
+ "import math\n",
+ "#initializing the variables:\n",
+ "Vi = -0.4;# in Volts\n",
+ "R1 = 4700;# in ohms\n",
+ "R2 = 10000;# in ohms\n",
+ "\n",
+ "#calculation:\n",
+ "A = 1 + (R2/R1)\n",
+ "Vo = A*Vi\n",
+ "\n",
+ "\n",
+ "#Results\n",
+ "print \"\\n\\n Result \\n\\n\"\n",
+ "print \"\\n(a) the voltage gain is \",round(A,2),\"\"\n",
+ "print \"\\n(b) output voltageis \",round(Vo,2),\" V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "\n",
+ " Result \n",
+ "\n",
+ "\n",
+ "\n",
+ "(a) the voltage gain is 3.13 \n",
+ "\n",
+ "(b) output voltageis -1.25 V"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 8, page no. 287</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#determine the output voltage, Vo\n",
+ "from __future__ import division\n",
+ "import math\n",
+ "#initializing the variables:\n",
+ "V1 = 0.5;# in Volts\n",
+ "V2 = 0.8;# in Volts\n",
+ "V3 = 1.2;# in Volts\n",
+ "R1 = 10000;# in ohms\n",
+ "R2 = 20000;# in ohms\n",
+ "R3 = 30000;# in ohms\n",
+ "Rf = 50000;# in ohms\n",
+ "\n",
+ "#calculation:\n",
+ "Vo = -1*Rf*(V1/R1 + V2/R2 + V3/R3)\n",
+ "\n",
+ "\n",
+ "#Results\n",
+ "print \"\\n\\n Result \\n\\n\"\n",
+ "print \"\\n output voltageis \",round(Vo,2),\" V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "\n",
+ " Result \n",
+ "\n",
+ "\n",
+ "\n",
+ " output voltageis -6.5 V"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 10, page no. 289</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "from __future__ import division\n",
+ "import math\n",
+ "from scipy import integrate\n",
+ "#initializing the variables:\n",
+ "Vs = -0.75;# in Volts\n",
+ "R = 200000;# in ohms\n",
+ "C = 2.5E-6;# in Farads\n",
+ "t = 0.1;# in secs\n",
+ "\n",
+ "#calculation:\n",
+ "f = lambda x,a : a*1\n",
+ "y, err = integrate.quad(f, 0, 0.1, args=(-0.75,))\n",
+ "Vo = (-1/(C*R))*y\n",
+ "\n",
+ "#Results\n",
+ "print \"\\n\\n Result \\n\\n\"\n",
+ "print \"\\n output voltage is \",Vo,\" V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "\n",
+ " Result \n",
+ "\n",
+ "\n",
+ "\n",
+ " output voltage is 0.15 V"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 11, page no. 290</h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Determine the output voltage Vo\n",
+ "from __future__ import division\n",
+ "import math\n",
+ "#initializing the variables:\n",
+ "V1a = 0.005;# in Volts\n",
+ "V2a = 0;# in Volts\n",
+ "V1b = 0;# in Volts\n",
+ "V2b = 0.005;# in Volts\n",
+ "V1c = 0.05;# in Volts\n",
+ "V2c = 0.025;# in Volts\n",
+ "V1d = 0.025;# in Volts\n",
+ "V2d = 0.05;# in Volts\n",
+ "R1 = 10000;# in ohms\n",
+ "R2 = 10000;# in ohms\n",
+ "R3 = 100000;# in ohms\n",
+ "Rf = 100000;# in ohms\n",
+ "\n",
+ "#calculation:\n",
+ "Vo1 = -1*Rf*V1a/R1\n",
+ "Vo2 = (R3/(R2+R3))*(1 + (Rf/R1))*V2b\n",
+ "Vo3 = -1*Rf*(V1c-V2c)/R1\n",
+ "Vo4 = (R3/(R2+R3))*(1 + (Rf/R1))*(V2d-V1d)\n",
+ "\n",
+ "\n",
+ "#Results\n",
+ "print \"\\n\\n Result \\n\\n\"\n",
+ "print \"\\n (a)output voltage is \",round(Vo1,2),\" V\"\n",
+ "print \"\\n (b)output voltage is \",round(Vo2,2),\" V\"\n",
+ "print \"\\n (c)output voltage is \",round(Vo3,2),\" V\"\n",
+ "print \"\\n (d)output voltage is \",round(Vo4,2),\" V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "\n",
+ " Result \n",
+ "\n",
+ "\n",
+ "\n",
+ " (a)output voltage is -0.05 V\n",
+ "\n",
+ " (b)output voltage is 0.05 V\n",
+ "\n",
+ " (c)output voltage is -0.25 V\n",
+ "\n",
+ " (d)output voltage is 0.25 V"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file |